The mol. mass odf NaCl determined by studying freezing point depression of its 0.5% aq. sol. is 30 . The degree of dissociation of NaCl will be?
Observed Molar Mass of NaCl = 30
Actual Molar Mass of NaCl = 58.5 g/mol
Van't Hoff Factor, i = Normal Molar Mass of NaCl / Observed Molar Mass of NaCl
i = 58.5 / 30 = 1.95
NaCl ionizes as :
NaCl ⇔ Na+ + Cl-
Let α be the degree of dissociation, then
Moles after dissociation
Total number of moles after dissociation = 1 - α + α + α = 1 + α
We know that , i = Observed moles of solute / Normal moles of solute
i = (1 + α) / 1
(1 + α) / 1 = 1.95
1 + α = 1.95
α = 1.95 – 1 = 0.95
Therefore, degree of dissociation of 0.5% aqueous solution of NaCl = 0.95 which means that at this concentration NaCl is 95% dissociated.